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A steel cable with cross-sectional area 3.00 cm\(^2\) has an elastic limit of 2.40 \(\times\) 10\(^8\) Pa. Find the maximum upward acceleration that can be given a 1200-kg elevator supported by the cable if the stress is not to exceed one-third of the elastic limit.

Short Answer

Expert verified
The maximum upward acceleration is 10.19 m/s².

Step by step solution

01

Understanding the Problem

We need to determine the maximum upward acceleration of a 1200-kg elevator supported by a steel cable. The stress on the cable should not exceed one-third of the cable's elastic limit. The cross-sectional area of the cable is 3.00 cm\(^2\), and the elastic limit is 2.40 \times 10^8 Pa.
02

Convert Units

First, convert the cross-sectional area from cm\(^2\) to m\(^2\). Since 1 cm\(^2\) = 0.0001 m\(^2\),\[3.00\,\text{cm}^2 = 3.00 \times 0.0001 \,\text{m}^2 = 3.00 \times 10^{-4} \,\text{m}^2\]
03

Calculate Allowed Stress

The stress on the cable should not exceed one-third of its elastic limit. Elastic limit is 2.40 \times 10^8 Pa. Thus,\[\text{Maximum Stress} = \frac{1}{3} \times 2.40 \times 10^8 \,\text{Pa} = 0.80 \times 10^8 \,\text{Pa}\]
04

Apply the Stress Formula

Stress \( \sigma \) is defined as force \( F \) per unit area \( A \):\[ \sigma = \frac{F}{A} \]Solving for the force,\[ F = \sigma \times A \]Substitute the values for maximum stress and area,\[ F = 0.80 \times 10^8 \,\text{Pa} \times 3.00 \times 10^{-4} \,\text{m}^2 = 24000 \,\text{N}\]
05

Determine Total Force Needed

The force \( F \) accounted for both gravitational force and upward acceleration. The gravitational force (weight) \( W \) for the elevator is:\[ W = m \cdot g = 1200 \,\text{kg} \times 9.81 \,\text{m/s}^2 = 11772 \,\text{N}\]
06

Calculate Net Force for Acceleration

The net force for acceleration (\( F_{\text{net}} \)) is the total force minus the weight of the elevator:\[ F_{\text{net}} = F - W = 24000 \,\text{N} - 11772 \,\text{N} = 12228 \,\text{N}\]
07

Find Maximum Acceleration

Using Newton's Second Law (\( F = m \cdot a \)), solve for acceleration \( a \):\[ a = \frac{F_{\text{net}}}{m} = \frac{12228 \,\text{N}}{1200 \,\text{kg}} = 10.19 \,\text{m/s}^2\]
08

Conclusion

The maximum upward acceleration that can be given to the elevator while keeping the stress within safe limits is 10.19 m/s\(^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stress and Strain
When we talk about physics, especially in materials science, stress and strain are critical concepts to understand. Stress is essentially the force exerted on a material per unit of cross-sectional area. It helps us know how much force a material can withstand before it starts to deform. Stress is measured in Pascals (Pa), which is equivalent to one Newton per square meter. For example, in the exercise, the steel cable can handle a certain stress limit, termed as its elastic limit, which is 2.40 \( \times \) 10e8 Pa.
Strain, on the other hand, is the measure of deformation that a material undergoes when the stress is applied. While stress refers to the cause, strain is the effect, expressed as the ratio of change in dimension to the original dimension. Importantly, strain is dimensionless, meaning it has no units.
  • Stress formula: \( \sigma = \frac{F}{A} \), where \( F \) is the force applied, and \( A \) is the cross-sectional area.
  • Relationship: Higher stress results in higher strain until the material's elastic limit is reached, beyond which permanent deformation may occur.
In summary, understanding stress and strain helps engineers design structures that can support intended loads without failing.
Newton's Laws of Motion
To solve problems involving motion like the one given, Newton's Laws of Motion provide the necessary framework. The second law is especially important here, as it relates the net force acting on an object to its mass and acceleration. Expressed mathematically, it is \( F = m \cdot a \), where \( F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration.
In our problem:
  • The gravitational force acting downward on the elevator is the product of its mass (1200 kg) and the gravitational acceleration (9.81 m/se2), resulting in a force of 11772 N.
  • The net force acting on the elevator is the total force the cable can exert, minus this gravitational force. This determines how much additional force can be used to accelerate the elevator upward.
By applying Newton's Second Law, we can determine the maximum upward acceleration that the cable and elevator system can safely achieve without surpassing the allowable stress.
Elasticity and Material Properties
Elasticity is a property that describes how a material returns to its original shape after the removal of a stress or force. This is crucial when evaluating material properties for use in construction or any application where a material must withstand forces without permanent deformation.
For materials like steel, there is a specific limit known as the elastic limit. This is the maximum stress level a material can endure before it permanently deforms. In our example, the steel cable has an elastic limit of 2.40 \( \times \) 10e8 Pa.
Elasticity is paramount in:
  • Designing structures that return to their original shape after being subjected to loads.
  • Selecting materials that will not fail or permanently deform under expected stress conditions.
The concept of elasticity assures engineers and architects that the materials they choose will maintain structural integrity under normal use. Understanding elasticity and material properties ensures that constructions are both safe and efficient, providing necessary longevity and reliability.

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Most popular questions from this chapter

A 3.00-m-long, 190-N, uniform rod at the zoo is held in a horizontal position by two ropes at its ends (\(\textbf{Fig. E11.19}\)). The left rope makes an angle of 150\(^\circ\) with the rod, and the right rope makes an angle \(\theta\) with the horizontal. A 90-N howler monkey (\(Alouattase\) \(niculus\)) hangs motionless 0.50 m from the right end of the rod as he carefully studies you. Calculate the tensions in the two ropes and the angle \(\theta\). First make a free-body diagram of the rod.

The compressive strength of our bones is important in everyday life. Young's modulus for bone is about 1.4 \(\times\) 10\(^{10}\) Pa. Bone can take only about a 1.0% change in its length before fracturing. (a) What is the maximum force that can be applied to a bone whose minimum cross-sectional area is 3.0 cm\(^2\)? (This is approximately the crosssectional area of a tibia, or shin bone, at its narrowest point.) (b) Estimate the maximum height from which a 70-kg man could jump and not fracture his tibia. Take the time between when he first touches the floor and when he has stopped to be 0.030 s, and assume that the stress on his two legs is distributed equally.

A vertical, solid steel post 25 cm in diameter and 2.50 m long is required to support a load of 8000 kg. You can ignore the weight of the post. What are (a) the stress in the post; (b) the strain in the post; and (c) the change in the post's length when the load is applied?

You are a summer intern for an architectural firm. An 8.00-m-long uniform steel rod is to be attached to a wall by a frictionless hinge at one end. The rod is to be held at 22.0\(^\circ\) below the horizontal by a light cable that is attached to the end of the rod opposite the hinge. The cable makes an angle of 30.0\(^\circ\) with the rod and is attached to the wall at a point above the hinge. The cable will break if its tension exceeds 650 N. (a) For what mass of the rod will the cable break? (b) If the rod has a mass that is 10.0 kg less than the value calculated in part (a), what are the magnitude and direction of the force that the hinge exerts on the rod?

A uniform metal bar that is 8.00 m long and has mass 30.0 kg is attached at one end to the side of a building by a frictionless hinge. The bar is held at an angle of 64.0\(^\circ\) above the horizontal by a thin, light cable that runs from the end of the bar opposite the hinge to a point on the wall that is above the hinge. The cable makes an angle of 37.0\(^\circ\) with the bar. Your mass is 65.0 kg. You grab the bar near the hinge and hang beneath it, with your hands close together and your feet off the ground. To impress your friends, you intend to shift your hands slowly toward the top end of the bar. (a) If the cable breaks when its tension exceeds 455 N, how far from the upper end of the bar are you when the cable breaks? (b) Just before the cable breaks, what are the magnitude and direction of the resultant force that the hinge exerts on the bar?

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